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Let $k$ be a commutative ring. Prove that a Lie $k$-algebra $\mathfrak{g} = 0$ iff $U\mathfrak{g} = k$. Use the adjoint representaion.

Here is my attempt at it:

The only non-trivial statement is that if $U\mathfrak{g} = k$, then $\mathfrak{g} = 0$.

There is an isomorphism between categories of $\mathfrak{g}$-modules and $U\mathfrak{g}$-modules. A module is the same as a representation. Consider $\operatorname{ad} \mathfrak{g}$ and apply the isomorphism, you'll get a representation $\varphi: U\mathfrak{g} = k \to \operatorname{End} \mathfrak{g}$ s.t. $\operatorname{im} \varphi \subset Z(\operatorname{End} \mathfrak{g})$. This implies that $\operatorname{ad} \mathfrak{g} = 0$, so $\mathfrak{g}$ is abelian.

However, if $\mathfrak{g}$ is abelian, then $U\mathfrak{g}$ is simply $S \mathfrak{g}$, the symmetric algebra over $\mathfrak{g}$, and since $\mathfrak{g} \subset S\mathfrak{g}$, it follows that $S \mathfrak{g} = k$ iff $\mathfrak{g} = 0$, Q.E.D.

Is this proof correct and complete? Have I used the most optimal way, or is there a more elegant way of using $\operatorname{ad}$ to prove this proposition? Also, if there's a more elegant way to prove this without using $\operatorname{ad}$, please do share :)

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This looks fine. – M Turgeon May 24 '12 at 11:48
If you want a quick proof, you can use the Poincaré-Birkhoff-Witt theorem, and then it is immediate -- but you need $k$ to be a field. – M Turgeon May 24 '12 at 11:51
@MTurgeon More precisely, it requires $\mathfrak{g}$ to be a free $k$-module. – Alexei Averchenko May 24 '12 at 12:08
@MTurgeon Actually, now I'm not sure that $\operatorname{im} \varphi \subset Z(\operatorname{End} \mathfrak{g})$. E.g. $\varphi(x) = \begin{pmatrix}x & 0 \\ 0 & 0\end{pmatrix}$ does not commute with everything, although of course $[\varphi(x), \varphi(y)] = 0$ for any $x$ and $y$. – Alexei Averchenko May 25 '12 at 5:43

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